# Electrolysis of AgNO₃: Mass of Silver Plated Let's find the mass of silver (Ag) plated on the cathode when a current of **1.2 A** is passed for **8 minutes** through AgNO₃ solution. --- ## Step 1: Calculate Total Charge Passed The total charge (\( Q \)) passed is: \[ Q = I \times t \] where - \( I = \) current in amperes (A) - \( t = \) time in seconds (s) Given: - \( I = 1.2\, \text{A} \) - \( t = 8\, \text{min} = 8 \times 60 = 480\, \text{s} \) So, \[ Q = 1.2 \times 480 = 576\, \text{C} \] --- ## Step 2: Write the Half-Reaction at the Cathode For AgNO₃, the reduction at the cathode is: \[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \] This means **1 mole of electrons plates 1 mole of silver**. --- ## Step 3: Calculate Moles of Electrons Passed Use Faraday’s constant (\( F = 96\,485\, \text{C mol}^{-1} \)): \[ \text{Moles of electrons} = \frac{Q}{F} = \frac{576}{96\,485} \approx .00597\, \text{mol} \] --- ## Step 4: Moles of Silver Plated From the half-reaction, **1 mole of e⁻ plates 1 mole of Ag**. So, \[ \text{Moles of Ag plated} = \text{Moles of electrons} = .00597\, \text{mol} \] --- ## Step 5: Calculate Mass of Silver Plated Molar mass of Ag = **107.87 g/mol**. \[ \text{Mass of Ag} = \text{Moles} \times \text{Molar mass} = .00597 \times 107.87 \approx .644\, \text{g} \] --- ## Final Answer **The mass of silver plated out on the cathode is:** \[ \boxed{.64\, \text{g}} \] (Rounded to two significant figures, matching the data precision)